Finding the closure of A

[thedarjeeling]

Let A = {1,2,3,4,5,6,7,8,9,10} and T = {R, empty, {1}, {1, 3}, {1,3,5}....}

What is the closure of A?

The definition we have for a point x being in the closure of A is that for all open sets containing x, the intersection of the open set and set A is nonempty. And then the closure of A is the collection of all points x in the closure.

So therefore I concluded that {1,3,5,7,9} is the closure of A, since it's obvious for each of those elements in A, that all the open sets containing them intersect A.

However I don't understand something, because we proved in class that the closure of any set is closed, but I don't see how this set {1,3,5,7,9} is closed. Its complement in R is obviously not open, because it contains all sorts of non natural numbers like irrationals, and if its complement is not open, then it can't be closed.

I must be looking at this wrong!

 

[pka]

Let A = {1,2,3,4,5,6,7,8,9,10} and T = {R, empty, {1}, {1, 3}, {1,3,5}....}
What is the closure of A?
The definition we have for a point x being in the closure of A is that for all open sets containing x, the intersection of the open set and set A is nonempty. And then the closure of A is the collection of all points x in the closure.
So therefore I concluded that {1,3,5,7,9} is the closure of A, since it's obvious for each of those elements in A, that all the open sets containing them intersect A.

I really don't understand the topology defined here.
But I think your concept of closure is off.
The statement that \(\displaystyle x\) is a point of \(\displaystyle \overline{A}\) means that \(\displaystyle x\in A\) or else is a limit point of \(\displaystyle A\).
Therefore \(\displaystyle A\subseteq\overline{A}~.\)

 

[thedarjeeling]

Yes, I realized my closure was off, I think I now understand it to be simply the real numbers for this topology, not the finite set I listed.

 

[HallsofIvy]

Then perhaps you should restate the problem. What, exactly, is X? What is A? And what sets are in T?

 

[thedarjeeling]

X = real numbers

A = collection of discrete points {1,2,3,4,5,6,7,8,9,10}

Tau = X, empty set, {1}, {1,3}, {1,3,5}, {1,3,5,7}, ......ad infinitum

So tau has the universe and the empty set (condition one met), any 2 arbitrary intersections is still in tau (condition 2 met), and any arbitrary number of unions is also in tau (condition met), so tau is a topology.

The problem really asked for 3 things, the interior, limit points, and closure of A.

The interior is obvious, because it is the largest open set that is contained within A. Therefore it's {1,3,5,7,9}

The limit points I realized is actually everything excluding 1. In other words, it's (-infinity, 1) union (1, infinity). I realized this because even though say for instance the number 0 is clearly not within A, however it satisfies the definition of a limit point of A, because for all open sets that contain 0 (there is only one open set which does so, which is R), the open set intersected with set A is nonempty and different than 0 (clearly, R intersected with A is nonempty). So even though 0 seems nonsensical because it's not in A, it still qualifies as a limit point of A. Same with every single number on the number line. The exception is 1, because the definition of limit points say the intersection of all open sets has to be nonempty and differing from the point the open set contains, but clearly {1} is an open set and its only intersection with A is {1} so that disqualifies it. The other open sets have at least a different point of intersection than the point itself.

The closure then is everything including 1, since there is no restriction on whether the intersection can be the point itself, so it's just the real numbers.